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I found this amazing question in the last calculus exam, but I don't know how to answer.

Let $T(x) = \ln(1+a) + \frac{1}{1+a}(x-a) - \frac{1}{2(1+a)^2}(x-a)^2 +...+ \frac{(-1)^{n+1}}{n(1+a)^n}(x-a)^n...$

be the expansion of Taylor series $T(x)$ of the function $f(x)=\ln(1+x)$ centered at $a>-1$

and $-1<x\le1+2a$ the interval of convergence of $T(x)$. In that interval $T(x)=f(x)$

1) How can I find all the values ​​of $a$ for calculating $\ln(2)$ with $T(x)$?

2) Furthermore, for $a=0$, how I can determine the degree $n$ of the $T_{n}(x)$ for estimating $\ln(2)$ with an error less than or equal to $1/4$? And give the estimate value of $\ln(2)$.

Any ideas where I can start?

Thanks in advance

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The natural choice for $a$ is $a=0$, since that makes all the coefficients easy to compute. Another plausible choice is $a=e-1$; this is closer to $2$ so the approximation will be better, but it's no longer possible to meaningfully calculate by hand. If for some reason you have a calculator with an "e" button but a broken "log" button, this is the way to go.

As for estimating error, with $a=0$ and $x=1$, you have an alternating series, so you can use the alternating series error estimate.

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  • $\begingroup$ I thought I needed to find a range for $a$. I don't need that? $\endgroup$ – user78723 May 23 '13 at 0:32
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In order to have $\ln(2)\approx T_n(2)$, you need $2$ to be within the interval of convergence. Since the interval goes only up to $1+2a$, this requires $1+2a\ge 2$. Solve for $a$.

vadim123 already answered part 2.

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